Answer: Least Common Multiplier of a6m, 0.003 moles 0.01elamos
Steps 6m, 0.003 moles
Compute an expression comprised of Factors that appear either in a6m or 0.003 moles
= 0.018elamos
Explanation:
According to this formula:
Q = M*C*ΔT
when we have M ( the mass of water) = 200 g
and C ( specific heat capacity ) of water = 4.18 J/gC
ΔT (the difference in temperature) = Tf - Ti
= 100 - 24
= 76°C
So by substitution:
Q = 200 g * 4.18 J/gC * 76 °C
= 63536 J
∴ the amount of heat which be added and absorbed to raise the temp from 24°C to 100°C is = 63536 J
Answer:
Answer:
<h2>1.75 x 10-4 M</h2>
Explanation:
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<u>Answer:</u> The activation energy for the reaction is 40.143 kJ/mol
<u>Explanation:</u>
To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:
![\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B317K%7D%7D%7BK_%7B278K%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 317 K = 
= equilibrium constant at 278 K = 
= Activation energy = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = 278 K
= final temperature = 317 K
Putting values in above equation, we get:
![\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B3.050%5Ctimes%2010%5E8%7D%7B3.600%5Ctimes%2010%5E%7B7%7D%7D%29%3D%5Cfrac%7BE_a%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B278%7D-%5Cfrac%7B1%7D%7B317%7D%5D%5C%5C%5C%5CE_a%3D40143.3J%2Fmol%3D40.143kJ%2Fmol)
Hence, the activation energy for the reaction is 40.143 kJ/mol
